$\mathrm{V}_{\text {outer }}=\frac{\mathrm{kQ}}{2 \mathrm{R}}$
$\mathrm{V}_{\text {inner }}=\frac{\mathrm{k} \mathrm{Q}}{\mathrm{R}}$
$\mathrm{V}_{\text {inner }}-\mathrm{V}_{\text {outer }}=\frac{\mathrm{kQ}}{2 \mathrm{R}}=\mathrm{V}$
After
$\mathrm{V}_{\text {inner }}=\mathrm{V}_{\text {outer }}=\frac{\mathrm{k} Q}{2 \mathrm{R}}=\mathrm{V}$
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($1$) Which one of the following statements is correct?
($A$) The balls will stick to the top plate and remain there
($B$) The balls will bounce back to the bottom plate carrying the same charge they went up with
($C$) The balls will bounce back to the bottom plate carrying the opposite charge they went up with
($D$) The balls will execute simple harmonic motion between the two plates
($2$) The average current in the steady state registered by the ammeter in the circuit will be
($A$) zero
($B$) proportional to the potential $V_0$
($C$) proportional to $V_0^{1 / 2}$
($D$) proportional to $V_0^2$
Give the answer quetion ($1$) and ($2$)